Question

A liquid of mass m and specific heat S is at temperature 0°C. Another liquid of mass 2m
and specific heat S/2 is at temperature 38°C. If these two liquids are mixed, the
equilibrium temperature is​

Answer 1

Answer:

Correct option is

B

t

Heat lost by one liquid=Heat gained by another liquid

m

1

.C

1

.ΔT

1

=m

2

.C

2

.ΔT

2

m(S)(2t−T)=m(1.5S)(T−

3

t

)

2t−T=3

2

T

−

2

t

5

2

T

=5

2

t

T=t

Answer 2

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$\large\bf{\underline{\red{VERIFIED✔}}}$

$θ= \dfrac{m_1s_1θ_1 + m_2s_2θ_2}{m_1s_1 + m_2s_2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = \frac{(m \times s \times t) + (\frac{m}{2} \times 2s \times 2t) }{(m \times s) + ( \frac{m}{2} \times 2s)} \\ \\ = > \frac{3mST}{2mS} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = > \frac{3}{2}T \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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